For which of the following curves, the line x + √3 y = 2√3 is the tangent at the point ( 3√3/2,1/2 )?

Solution

Maths sir · 4 steps

The point lies on all four curves, so write the tangent at (x₁, y₁) for each and see which one matches.

1. Point on every curve. With x₁ = 3√3/2 and y₁ = 1/2 we get x₁² = 27/4 and y₁² = 1/4. Substituting shows that the point satisfies all four equations, so being on the curve does not decide anything.
2. Tangent rule. For a second-degree curve, the tangent at (x₁, y₁) is T = 0. Replace x² by xx₁, y² by yy₁, and x by (x + x₁)/2. Then compare the result with x + √3 y = 2√3, whose slope is -1/√3.

   Curve	Tangent at the point	Simplified	Slope
   (a) 2x² - 18y² = 9	3√3 x - 9y = 9	√3 x - 3y = 3	1/√3
   (b) y² = x/6√3	y/2 = (x + x₁)/12√3	6√3 y = x + 3√3/2	1/6√3
   (c) x² + 9y² = 9	3√3/2x + 9/2y = 9	x + √3 y = 2√3	-1/√3
   (d) x² + y² = 7	3√3/2x + 1/2y = 7	3√3 x + y = 14	-3√3

3. Check (c). Multiply 3√3/2x + 9/2y = 9 by 2/3√3. This gives x + 3/√3y = 6/√3, which is x + √3 y = 2√3, exactly the given line.
4. So. The line is the tangent to the ellipse x² + 9y² = 9 at the given point. The answer is (c).

The idea to keep. A line passing through a point of a curve is not automatically the tangent there. Write T = 0 at that point and compare the result with the given line. Matching slopes is the quickest check.

Answer from the NTA final answer key. Solution by PrepWiser.